Question

The single-slit diffraction pattern shown in the figure was produced with light of wavelength \(495 \mathrm{nm}\). The screen on which the pattern was projected is located a distance of \(2.77 \mathrm{~m}\) from the slit. The slit has a width of \(2.73 \mathrm{~mm}\). What is the width \(w\) of the central maximum?

Short answer

Answer: The width of the central maximum is approximately 26.8 mm.

Step-by-step solution

Find the angular width of the central maximum

The formula to find the angular width of the central maximum in a single-slit diffraction pattern is given by: \(sin(\theta) = \frac{\lambda}{a}\), where \(\theta\) is the angular width of the central maximum, \(\lambda\) is the wavelength and \(a\) is the slit width. In our problem, we have \(\lambda=495 \mathrm{nm}\) and \(a=2.73 \mathrm{mm}\). Firstly, let's convert the units to the same unit i.e., meters (m). We know that \(1 \mathrm{nm} = 10^{-9} \mathrm{m}\) and \(1 \mathrm{mm} = 10^{-3} \mathrm{m}\). Therefore, we have \(\lambda = 495\times10^{-9} \mathrm{m}\) and \(a = 2.73\times10^{-3} \mathrm{m}\). Now, plug these values into the formula for \(sin(\theta)\): \(sin(\theta) = \frac{495\times10^{-9} \mathrm{m}}{2.73\times10^{-3} \mathrm{m}}\).

Calculate the angular width of the central maximum

Now, to find the angular width, solve for \(\theta\): \(\theta = arcsin(\frac{495\times10^{-9} \mathrm{m}}{2.73\times10^{-3} \mathrm{m}})\). Use a calculator to find the value of \(\theta\): \(\theta = 0.004841\) radians (approximately). Note that this is the angular width for only one side of the central maximum. Since the central maximum is symmetrical, we need to double this value to find the total angular width: \(2\theta = 2 \times 0.004841 = 0.009682\) radians (approximately).

Find the linear width of the central maximum

We are given the distance between the slit and the screen as \(2.77 \mathrm{m}\). To find the linear width of the central maximum, we will use the formula for arc length, which is given by: \(w = L \times 2\theta\), where \(w\) is the linear width of the central maximum, \(L\) is the distance between the screen and the slit, and \(2\theta\) is the total angular width. Insert the given values into the formula: \(w = 2.77 \mathrm{m} \times 0.009682\).

Calculate the linear width of the central maximum

Calculate the value of w: \(w = 2.77 \mathrm{m} \times 0.009682 = 0.0268 \,\mathrm{m} \, (approx.)\) Finally, convert the width into millimeters for a more convenient unit: \(w = 0.0268 \,\mathrm{m} \times \frac{1000\, \mathrm{mm}}{1\, \mathrm{m}} = 26.8 \,\mathrm{mm}\) So, the width \(w\) of the central maximum of the single-slit diffraction pattern is approximately \(26.8 \,\mathrm{mm}\).